# Tag Info

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Electrons move because they are in a region of space with a non-zero electric field. They don't accelerate to high speed in a wire because they keep bumping into things; a kind of friction which dissipates energy much like the friction you are used to that explains why resistors get hot. In effect their speed depends on the strength of the local electric ...

5

Wouldn't this inductor's emf counteract the discharging capacitor and actually charge it? / stop the capacitor from fully discharging? The inductor doesn't care about what the charge state of the capacitor is. All it cares about is how quickly the current through it is changing, and it generates a back-voltage according to the equation V=L*dI/dt. You ...

4

When a capacitor of capacitance C is charged to a voltage V, and discharged through a resistor R, then the current will decay exponentially: $$I = I_0 e^{-t/RC}$$ The voltage on the capacitor will follow the same exponential decay, $$V = V_0 e^{-t/RC}$$ To answer your question one would have to make some assumptions. You will have to do the calculation ...

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Just a simple answer - there's nothing to dissipate energy. If there were a resistor in the circuit, it would dissipate energy as heat. Inductors and capacitors don't dissipate energy. The energy just sloshes back and forth between being stored in the magnetic field, and being stored in the electric field. It's just like a spring-mass system, where energy ...

3

Any wire circuit will have inductance and capacitance between the "outbound" and "return" wires - this immediately follows from very basic laws of physics, and in fact is intimately related to the finite propagation velocity of the electrical signal. The expression $$u=\frac{1}{\sqrt{LC}}$$ would give an infinite velocity if either $L$ or $C$ was zero... ...

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Notice We know that the reactance of a capacitor is given as $$\color{blue}{R_c=\frac{1}{2\pi fC}}$$ Where, $C$ is electric capacitance & $f$ is the frequency of source For an A.C. source, frequency, $\color{red}{f>0}\implies \color{blue}{R_c=\frac{1}{2\pi fC}>0}$ which means that a capacitor offers a constant resistance in A.C. circuit i.e. it ...

2

BEWARE THE SANDWICHES!!! :) In the spirit of math-avoidance sandwich-juggling, here's a better analogy, a visible one. The movable charges within conductive circuits are like silver bead-chains, like those little chains which attach the pens to desks in old-school banks. (Growing up I always played with these when mom was in the teller line. Do those ...

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The second paragraph from this IEEE reference follows: Every electrical engineer learns early of the two Kirchhoff laws, but not very many realize that they were published while he was still a student. The publication is {(vom Studiosus) Kirchhoff, “Ueber den Durchgang eines elektrischen Stromes durch eine Ebene, insbesonere durch eine kreisförmige,” ...

2

Let's think about the circuit you drew. It contains a battery, switch, bulb, and a very large inductor. In fact, the inductance of a wire that goes 10 times around the Earth can be calculated (I am going to assume an air core - in fact there is a piece of iron in the middle of the Earth which makes the resulting inductance greater). L\approx N^2 R \mu_0 ... 2 My argument was that because the resistance is higher, there must be less voltage going through at that point. This is probably the cause of the confusion. In spite of the usual formulation V=IR, in an electrical circuit Voltage and Resistance are the "inputs" to the equation and Current is the result or output. As an analogy, think of Newton's 2nd ... 2 R(V,I) = \frac{V}{I} by definition, it is not a gradient. r = \frac{dV}{dI} is called the fractional, differential, dynamical or small-signal resistance. It just happens that for resistors R(V,I) = R_0 is a constant, thus the two quantities are the same: r = R_0. 2 Using the "drop-in-a-bucket" trick, we find that an LC laddar has impedance \begin{align} Z &= Z_L + Z_C||Z \\ &= Z_L + \frac{Z_C Z}{Z + Z_C} \\ Z^2 - Z Z_L - Z_L Z_C &= 0 \\ Z &=\frac{1}{2} \left( Z_L \pm \sqrt{Z_L^2 + 4 Z_L Z_C} \right) \, . \end{align} The impedance of an inductance L is Z_L = i \omega L and the impedance of a ... 2 The inductor never creates a current in the opposite direction. An inductor creates an EMF to counteract the changing B field(Lenz law). The B field is changing because the current in the inductor is changing. So effectively, the inductor resists changes in current. So initially, the capacitor tries to discharge strongly but is slowed down by the ... 1 A resistor is defined as the circuit element for which the voltage across is proportional to the current through and the constant of proportionality is the resistance R:V_R = R\cdot I_R $$Clearly, for this linear relationship, it is also true that$$\frac{dV_R}{dI_R} = R However, for general circuit elements, the derivative of $V(I)$ is not a ...

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"Current takes the path of least resistance" is just a phrase people say but it's not totally accurate. When one path through the circuit has 0 resistance (a short), it is true that current follows that path only. It isn't true when you have multiple paths, with nonzero resistance, though. A better way of saying it would be "current flows through all paths ...

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The symmetry is that the lamps in the top half are the same as the lamps in the bottom half, albeit with a left/right switch. When S2 is closed, it is clear that the left/right placing makes no difference, since a and b are at the same voltage.

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Two questions: How can the ammeter tell how much current is flowing the resistor? since it's "behind" the resistor? There at least several means that current can be measured using different technologies. The early ammeters used galvanometric technology where a coil in the galvanometer becomes part of the current path. The coil generates a magnetic ...

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Why and how does a resistor limit the current flowing through the entire circuit? doesn't it limit only the current that is flowing past and after the resistor? First, this is a DC circuit (ignoring the switch) which is to say that the circuit voltages and currents are constant with time. Since that is the case, by conservation of electric charge, ...

1

9v batteries are made from 6 small 1.5v cells connected in series. So it has a very high internal resistance which limits how much current you can draw from one (and why they aren't used in high power devices) The resistance of your tongue is complicated, depends on how wet it is and the resistance changes as the current causes chemical changes. So you ...

1

1 amp current means 1 coloumb charge per second flows through the circuit. (1/1.6*10^-19) gives the electrons flowing through a point /second The above value * the total time (3600 seconds ) will give you the answer.

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(1) The voltage across a capacitor is proportional to the charge $Q$ (where it is understood that there is $+Q$ charge on one plate and $-Q$ charge on the other plate). (2) The voltage across the capacitor is initially zero so the $Q$ is initially zero. (3) The voltage across the battery is not zero so, if the battery and capacitor are connected together, ...

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Steady state means, in this context, ignoring transients due to initial conditions. For a capacitor, the steady state current due to a DC voltage source is zero. The steady state current due to an AC source is simply the (constant) amplitude of the sinusoidal current. To be clear, the current is changing in time and so is not constant. But, the amplitude ...

1

It sounds as though you're on the right track: if I understand correctly, you're saying that once you have found the voltage across the pair $R_1 and R_4$ (equal to the voltage across $R_2$ and $R_3$), presumably by lumping $R_1,\,R_2\,R_3\,R_4$ together through parallel addition of $R_1 +R_4$ and $R_2+R_3$, then you simply work out $V_o$ thinking of ...

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Immediately. They start to flow immediately. When you connect a resistor to the negative terminal of a battery and a wire to the positive terminal of the battery the whole resistor gets to lower potential and the whole wire gets to a higher potential. So when you start to connect that wire to the resistor you literally bring a positive voltage wire towards ...

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As you may know, it takes infinite time to charge a capacitor. So, the time when the capacitor is 100% charged never comes. Thus, we require a Time Constant to help us understand the time when the capacitor has got a decent amount of charge and after which the rate of charging becomes really slow and thus charging further is not of much use. You may also ...

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Just adding $0.02 for clarity: The formulas$I^2R$and$V^2/R$describe the power dissipated in a resistor. If you're considering the power lost in power transmission lines, the "resistor" is the power line, not the appliance in the home where the power is wanted. The$V$in the$V^2/R\$ formula is the voltage beetween the two ends of the resistor so, in ...

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The question is ill-posed; the electrons "know" nothing, and voltage is not a property of the electron (other than e.g. charge, which is a property). In fact, voltage is a pretty abstract concept; it is energy divided by charge. And that means explaining an abstract term by another abstract term. Let's be more fundamental: nature shows that charges exert ...

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Capacitance is about stored charge - more electrons flowing into something than flow out. This can happen in a piece of wire, although it can take a large amount of applied voltage to accumulate a small amount of excess electrons. In other words, a simple piece of wire has very low capacitance. Even a straight piece of wire will have inductance because any ...

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Perhaps this is what you are looking for: Screen capture: http://www.falstad.com/circuit/ The default circuit, as shown, is an LRC circuit. On the Schematic: Gray is 0V Green is Positive Voltage Red is Negative Voltage The yellow dots are a visualization of current: positive holes. The graphs along the bottom, from left to right, are for the ...

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